Friday, February 24

Noorul Huda

Aromaticity- A Brief Discussion

Whenever we talk about aromatic compounds, the first thing which comes in our mind is BENZENE. Yes ofcourse being a chemistry student, the first aromatic compund which we come across is benzene only. I still remember when i was in class 10th and my teacher taught us about structure of benzene I was so amused with the beauty of it's structure. I mean how can a molecule look so beautiful with conjugate double bonds and a cute hexagonal shape. Well I'm sure most of us must have thought the same way.



Okay so without wasting time lets come upon our topic of Aromaticity. Aromaticity was basically derived from the term Aroma [which means sweet smell], so initially it was considered that aromatic compounds are all those compounds having a pleasant smell.
For a chemist, an aromatic compound must have following properties:
  1. have an extremely high resonance energy (36 kcal/mol for benzene)
  2. undergo substitution rather than addition reactions
  3. have delocalized pi-electrons
So the Question arises how can we conclude whether a molecule is aromatic or not?

I've tried to make this simple for you,
here's the answer; For a molecule to be regarded as Aromatic, it must meet following 4 conditions. If any of these conditions are violated, no aromaticity is possible.

Condition 1: The Molecule Must Be Cyclic

 It's very simple if there's a Ring move to condition 2. If there’s no ring, forget it.

for e.g.: (Z)-1,3,5 hexatriene has the same number of pi bonds (and pi electrons) as benzene, but isn’t aromatic. No ring, no aromaticity.


Condition 2: Every atom in the ring must be conjugated

In order for aromaticty to exist, there must also be a continuous ring of p-orbitals around the ring that build up into a larger cyclic “pi system”.

One way of saying this is that every atom around the ring must be capable of conjugation with each other.

Remember that the “available p orbital” condition applies not just to atoms that are part of a pi bond, but also atoms bearing a lone pair, a radical, or an empty p orbital (e.g. Carbocation)

The key thing  that “kills” conjugation is an sp3 hybridized atom with four bonds to atoms. Such an atom cannot participate in resonance.



Condition 3: The Molecule Must Have [4n+2] Pi Electrons {The Huckel Rule}

The third condition is that the cyclic, conjugated molecule must have [4n+2] pi electrons. Benzene and cyclooctatetraene are both cyclic and conjugated, but benzene is aromatic and cyclooctatetraene is not. The difference is that benzene has 6 pi electrons, and cyclooctatetraene has 8.


However, this term [4n+2] causes a lot of confusion in organic chemistry. Many students stare at a molecule and try to figure out what “n” is.
“n” is not a property of the molecule!

“4n+2 is not a formula that you apply to see if your molecule is aromatic. It is a formula that tells you what numbers are in the magic series. If your pi electron value matches any number in this series then you have the capacity for aromaticity.”

The “magic series” is: 2, 6, 10, 14, 18, 22….. (and so on).
[4n+2] is just a mathematical shorthand for writing out the series [2, 6, 10, 14, 18, 22…] .

The numbers in this “magic series” are sometimes referred to as “Hückel Numbers” after Erich Hückel, who proposed this rule back in 1931.

The condition that aromatic molecules must have [4n+2] pi electrons is sometimes called “Hückel’s rule”.


In the figure below, molecules which fulfil Hückel’s rule are in green; those which do not fulfill Hückel’s rule are in red.

Note that we can count electrons in pi bonds as well as electrons from lone pairs (so long as the carbon isn’t already participating in a pi bond). So the cyclopentadiene anion has six pi electrons – 4 from the two double bonds, and two from the lone pair on carbon.

Here Question Arises-Which Electrons Count As “Pi Electrons”?

That seems easy. However, complications may arise when we have atoms in the ring which both participate in pi bonding and also have a lone pair. For example, pyridine, pyrrole etc.

Some Examples With 5-Membered Rings

Some molecules with five-membered rings can also present ambiguities.
The cyclopentadiene anion (below)  has a lone pair on one of the carbons. Can this lone pair contribute to the pi system?
Since that carbon is not involved in any pi-bonding, the answer is yes.
The total number of pi electrons for the cyclopentadiene anion equals 2 (from the lone pair) plus the 4 electrons in the two pi bonds, giving us a total of 6. This is a Hückel number and the cyclopentadiene anion is in fact aromatic.

A similar situation arises for pyrrole. The nitrogen bears a lone pair but  is not involved in a pi bond (unlike pyridine, above). Therefore it can contribute to the pi system and this gives us a total of 6 pi electrons once we account for the 4 electrons from the two pi bonds.

A curious case is furan, where the oxygen bears two lone pairs. Does this mean that furan has 8 pi electrons? No! 
Why not? Because as we noted above, each atom can contribute a maximum of one p-orbital towards the pi system. In furan, one lone pair is in a p orbital, contributing to the pi system; the other is in the plane of the ring. This gives us a total of 6 pi electrons. Furan is aromatic. (So is thiophene, the sulfur analog of furan).

Finally there is imidazole, which has two nitrogens. One nitrogen (the N-H) is not involved in a pi bond, and thus can contribute a full lone pair; the other is involved in a pi bond, and the lone pair is in the plane of the ring. This also gives us a total of 6 pi electrons once we account for the two pi bonds.


Condition 4: The Molecule Must Be Flat

The fourth condition for aromaticity is that the molecule must be planar.

   As with certain vertebrates, the only thing that preventing a molecule that fulfills the first three conditions from being flat is if the flat conformation is incredibly strained.
One example in this category is the molecule known as [10]-annulene, an isomer of which is drawn below left. In the trans, cistrans, cis, cis isomer, the molecule is cyclic, conjugated, and has 10 pi electrons, but the two marked hydrogens bump into one another when attempting to adopt a flat conformation.
The molecule is prevented from adopting planarity due to this punitive Van Der Waals strain , and is therefore not aromatic. 
Interestingly, if the hydrogens are removed and replaced with a bridging CH2 group, the strain is relieved and the pi bonds can adopt a planar conformation. The molecule below right shows the expected properties of an aromatic molecule.



 References: 

  1. Wikipedia
  2. Google image search
  3. masterorganic chemistry

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Thursday, February 23

Noorul Huda

CSIR Sample Question Paper

This post includes CSIR SAMPLE QUESTION PAPER for Chemical Sciences according to New Revised Scheme.


This Test Booklet will contain 120 (20 Part ‘A’+ 40 Part ‘B’ + 60 Part ‘C’) Multiple Choice Questions (MCQs).
Candidates will be required to answer 15 in part ‘A’, 35 in Part ‘B’ and 25 questions in Parts ‘C’ respectively.
In case any candidate answers more than 15, 35 and 25 questions in Parts A, B and C respectively only first 15, 35 and 25 questions in Parts A, B and C respectively will be evaluated.
Questions in Parts ‘A’ and ‘B’ carry two marks each and Part ‘C’ questions carry four marks each.
There will be negative marking @ 25% for each wrong answer. Below each question, four alternatives or responses are given.  Only one of these alternatives is the CORRECT answer to the question.


Also Read: CSIR New Revised Exam Scheme

Must Read: Top PhD Level Examination In India


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Monday, February 20

Noorul Huda

IIT JAM 2017 Question Paper with Answers

This Post includes IIT JAM Question Paper for the exam which held on 12th Feb 2017. The post also contains expected answers for each question.


A seperate Answer Key is provided for all Questions of Section A,
For section B and C answers are provided alongwith each question.

A Sample Image of the post is provided below. You can download full question paper in PDF from the link provided.

Click Here to Download Full Question paper in PDF Format 
Click Here to Download Answer Key

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Sunday, February 12

Noorul Huda

CSIR JUNE 2017

Joint CSIR-UGC Test for Junior Research Fellowship and Eligibility for Lectureship Tentatively on 18th June,2017.

CSIR will hold the Joint CSIR-UGC Test Tentatively on Sunday 18th June  2017 for determining the eligibility of the Indian National Candidates for the award of Junior Research Fellowships (JRF) NET and for determining eligibility for appointment of Lecturers in certain subjects areas falling under the faculty of science and Technology. The award of JRF to the successful eligible candidates will depend upon their finding admission/placement in a university/national laboratory/ institution of higher learning and research , as applicable.
Candidates with M.Sc qualification or under M.Sc Result Awaited category shall be eligible for Lectureship subject to fulfilling the eligibility criteria as laid down by the UGC.

Subjects of  CSIR UGC NET

The test includes the subjects –
(i)Chemical Sciences
(ii)Earth, Atmospheric, Ocean and Planetary Sciences
(iii) Life Sciences
(iv)Mathematical Sciences
(v)Physical Sciences

Education Qualification

M.Sc or Equivalent degree/Integrated BS-MS/BS-4 years /BE/BTECH/B PHARMA/MBBS  with atleast 55%  marks for general and OBC-Non Creamy layer  and 50% for SC,ST, persons with disability(PWD) Candidates.
Candidates enrolled for M.Sc or having completed 10+2+3 years of above qualifying examination as on 01.03.2016 are also eligible to apply in the above subject under Result awaited category on condition that they complete the qualifying degree with requisite percentage of marks within the validity period of 2 years. Such candidates will have to submit the attestation format duly certified by Head of the Department  from where candidate is appearing
BSc(Hons) or Equivalent degree or students enrolled in Integrated MS-PhD program with atleast 55%  marks for general and OBC-Non Creamy layer  and 50% for SC,ST, persons with disability(PWD) Candidates are also eligible to apply. Candidates with bachelor’s degree will be eligible for CSIR fellowship only after getting registered or enrolled for PhD/integrated PhD progarmme with the validity period of 2 years.
PhD degree holders who have passed Master’s degree prior to 19th sept 1991,with at least 50%marks are eligible to apply for lectureship only.

Must Visit: CSIR NEW PATTERN (Revised Exam Scheme)

AGE LIMIT & RELAXATION

For LS(NET) : No upper age limit
For JRF(NET) : Maximum 28 years as on 01-01-2017(upper age limit may be relaxed up to 5 years in case of candidates belonging to SC/ST/OBC (Non-creamy layer )/PwD and female candidates)

Important Dates

Date of single MCQ Examination                               Tentatively on 18.06.2017
Schedule for on-line Application (i) Start of on-line submission of Application and fee deposit through Bank Challan.
(ii) Date of close of deposit of fee
(iii) Date of close of on-line Application
08.02.2017
07.03.2017
08.03.2017
Last date of receipt of duly completed hard copy of on-line Application in Examination unit Last date of receipt of duly completed hard copy of on-line Application in Examination unit(from remote areas) 14.03.2017

21.03.2017
Last date of receipt of written request for change of examination centre only on merit basis 20.04.2017
Tentative date of publication of list of candidates registered for test on CSIR,HRDG website 16.05.2017
Issue of e-Admission certificate to registered candidates 1st week of june 2017

Must Read: 10 Best Tips to Crack CSIR-NET JRF (Preparation Guide)

CSIR UGC NET Application Form Fee

Candidates will have to deposit the examination fee at any branch of the Indian Bank all over the country. The amount of fee is:

Amount to pay Category type
1000/- General category candidates
500/- OBC-NCL category candidates
250/- SC/ST/ Physically Handicapped (PH) or Visually Handicapped (VH ) candidates
http://59.163.216.82:8080/jrf/online/index.jsp


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Tuesday, February 7

Noorul Huda

GATE 2017 Question Paper

This Post includes GATE Chemistry 2017 Question Paper which held on 5th Feb 2017. It is complete hardwork of our team mates and friends who helped us to compile various questions of the examination. We want to give special thanks to our contributors for this post  Mr.Suresh Kumar Pandey, Mr.Rajneesh Kumar, Miss.Siddhi Jaiswal, Mohd. Salman and all our group members.

We are unable to get all questions because some of them are data based and its quite difficult to collect them yet we request all our visitors if you know any other question beyond these questions then please post them on our Facebook Page,




Also Visit: GATE CHEMISTRY CY PREVIOUS YEAR QUESTION PAPERS WITH ANSWER KEYS





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Saturday, February 4

Noorul Huda

CSIR NEW EXAM PATTERN (Revised Exam Scheme)

From last few years CSIR was planning a change in its Entrance Exam Pattern to reduce competition and make examination a little bit more tough. On Feb 3rd 2017 a new Scheme is Published on CSIR website which suggests a few changes in previous exam pattern.


This post explains Major Changes and the things that are Not Changed in New Scheme.

The Screenshot of Official Notification is attached below:




 Things Not Changed in New Exam Scheme:

  1. Total Marks of Examination is still 200.
  2. Total Time Allowed for exam is still 3 hours.
  3. Part A of Question Paper still contains 20 Questions, out of which 15 are to be answered.
  4. As it was earlier All Questions are Multiple Choice Questions (MCQs).
  5. Syllabus of Exam remains SAME.
  6. Negative Marking remains same i.e 25% of total marks of question.


Major Changes in Exam Pattern are:

  • Total Number of Questions in Part B is reduced to 40, out of which any 35 are to be answered. (earlier it was total of 50 questions out of which any 35 were to be answered)
  • Total Number of Questions in Part C is reduced to 60, out of which any 25 are to be answered. (earlier it was total of 75 questions out of which 25 were to be answered)

 
 Click Here to Download Syllabus Chemical Science

 Wish You all the best for upcoming exams !!!


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Friday, February 3

Noorul Huda

GATE Chemistry Previous Year Question Paper with Explained Solution

GATE Chemistry CY previous year Question Paper with Explained solution can be downloaded from this page and can be utilized in best possible way to clear the GATE paper. Previous Year papers with Explained solution of the year 2015 and  2014 can be downloaded from the links below. Focus on tghe type of Question in each section so that it becomes easier to study different parts.





GATE exam is a career oriented examination in which the candidates sit with new hopes and have mindset in a proper way. Download the papers from below and practice more and more to be in flow of GATE paper.

Visiters Also Downloaded: GATE Chemistry Practice Problem Sets

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